WEBVTT
00:00:00.760 --> 00:00:13.300
Given that vector 𝐀 is equal to negative two, three and vector 𝐁 is equal to negative one, eight, find the dot product of 𝐀 minus 𝐁 and 𝐁.
00:00:14.830 --> 00:00:19.740
We will begin this question by subtracting vector 𝐁 from vector 𝐀.
00:00:20.660 --> 00:00:27.450
When subtracting vectors, we simply need to subtract their individual corresponding components.
00:00:28.350 --> 00:00:33.040
Subtracting negative one from negative two gives us negative one.
00:00:33.680 --> 00:00:37.680
This is because this is the same as adding one to negative two.
00:00:38.270 --> 00:00:42.020
Subtracting eight from three gives us negative five.
00:00:43.260 --> 00:00:49.110
The vector 𝐀 minus 𝐁 is therefore equal to negative one, negative five.
00:00:49.920 --> 00:01:01.750
We recall that in order to find the dot or scalar product of two vectors, we begin by finding the product of their individual components and then find the sum of these values.
00:01:02.420 --> 00:01:09.970
In this question, we need to find the dot product of negative one, negative five and negative one, eight.
00:01:10.790 --> 00:01:14.680
Both of the 𝑥-components of these vectors are negative one.
00:01:15.440 --> 00:01:18.950
The 𝑦-components are negative five and eight.
00:01:19.700 --> 00:01:28.440
Multiplying two negative numbers gives us a positive answer, whereas multiplying a negative number by a positive number gives a negative answer.
00:01:29.650 --> 00:01:32.880
This leaves us with one minus 40.
00:01:33.480 --> 00:01:38.980
Taking 40 away from one gives us an answer of negative 39.
00:01:39.760 --> 00:01:43.700
This is the dot product of 𝐀 minus 𝐁 and 𝐁.